Solution:
Solution:
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Question111
Let 1
16,aandbbeinG.P.and 1
a,1
b,6bein(a)P.,wherea,b>0.Then,
72(a+b)isequalto.........
[2021,16MarchShift-II]
Answer:14
Solution:
Solution:
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Question112
Ifα,βarenaturalnumbers,suchthat100α−199β = (100)(100) + (99)(101)+
(98)(102) + ... + (1)(199),thentheslopeofthelinepassingthrough(α,β)
andoriginis
[2021,18MarchShift-1]
Options:
Given,set{11,8,21,16,26,32,4}
Byobservation,wecansaythat
AP = {11,16,21,26, ...}
GP = {4,8,16,32, ...}
5m +6=4.2n−1
5m +6=2n+1
So,(2n+1−6)shouldbeamultipleof5.Theunitdigitof2kis2,4,6,8.So,when6issubtractedfrom2n+1,thepossibleunit
digitswillbe6,8,0,2.Only0isdivisibleby5.Hence,2n+1unitdigithastobe6.
2n+1=24,28,212,216...
As,216willnotbea4digitnumber,so,commonterms = {16,256,4096}
∴Numberofcommonterms = 3
Given, GP =1
16,a,b
⇒a2=b
16
andgiven,AP =1∕a,1∕b,6
⇒2
b=1
a+6
⇒2
16a2=1
a+6
⇒1
8a2=1+6a
a
⇒1=8a(1+6a)
⇒48a2+8a −1=0
⇒ (4a +1)(12a −1) = 0
⇒a= −1∕4 or 1 ∕12
Asperthequestion,a>0
∴a=1∕12
b=16a2=16⋅1
144 =1
9
∴72(a+b) = 72 1
12 +1
9
=6+8
=14
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